LA 6476 Outpost Navigation (DFS+剪枝)

题目链接

Solution

  DFS+剪枝

  对于一个走过点k,如果有必要再走一次,那么一定是走过k后在k点的最大弹药数增加了.否则一定没有必要再走.

记录经过每个点的最大弹药数,对dfs进行剪枝.

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <map>
using namespace std;
map<string, int> pos;
struct Edge {
int v, c, next;
} edge[];
int head[], cnt;
int vis[], amm[], aim[],sum[];
int cs, n, m, ans;
string s, t; void addedge (int u, int v, int c) {
edge[++cnt].v = v, edge[cnt].c = c;
edge[cnt].next = head[u];
head[u] = cnt;
}
void dfs (int x, int s, int k) {
if(k>=ans) return;
if (aim[x]) {
ans = min (ans, k);
return ;
}
if (!vis[x]) s += amm[x], vis[x] = ;
if(vis[x]&&sum[x]>=s) return;
sum[x]=max(sum[x],s);
for (int i = head[x]; i; i = edge[i].next) {
int v = edge[i].v, c = edge[i].c;
if (s >= c) dfs (v, s - c, k + c);
}
vis[x] = ;
}
int main() {
scanf ("%d", &cs);
while (cs--) {
pos.clear();
memset (head, , sizeof head);
memset(sum,,sizeof sum);
memset(vis,,sizeof vis);
cnt = , ans = 0x7fffffff;
scanf ("%d %d", &n, &m);
for (int i = ; i <= n; i++) {
cin >> s;
pos[s] = i;
cin >> amm[i] >> s;
if (s[] == 'y') aim[i] = ;
else aim[i] = ;
}
for (int i = , x; i <= m; i++) {
cin >> s >> t >> x;
int u = pos[s], v = pos[t];
addedge (u, v, x);
addedge (v, u, x);
}
dfs (, , );
if (ans != 0x7fffffff) printf ("%d\n", ans);
else puts ("No safe path");
}
}
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